# $f(A) \cap f(B) = \emptyset \, \forall \, A,B \subseteq S,$ and $A \cap B \, = \emptyset \implies f(A-B)=f(A)-f(B)$ for $B \subseteq A$

This is the full question:

$$f: S \to T$$ is a function, then the following statements are equivalent:

a) $$f$$ is one-one on $$S$$

b) $$f(A \cap B) = f(A)\cap f(B) \, \forall \, A,B \subseteq S$$

c) $$f^{-1}(f(A))=A \hspace{4mm} \forall \, A\subseteq S$$

d) $$f(A) \cap f(B) = \emptyset \, \forall \, A,B \subseteq S,$$ and $$A \cap B \, = \emptyset$$

e) $$f(A-B)=f(A)-f(B) \,\, \forall \, A,B \subseteq S$$ and $$B \subseteq A$$

I am having trouble is understanding a step in the proof of $$\, d \implies e$$ , Here is the proof

If $$B \subseteq A \text{ then } (A-B)\cap B = \emptyset \\ \implies f(A-B) \cap f(B) = \emptyset \\ \implies f(A-B) \subseteq f(A)-f(B) \hspace{5mm} (*) \\ \text{if } \, y \in f(A-B) \text{ then } y=f(x) \text{ for some } x \in A \text{ and } x \notin B \\ x\in A-B \implies f(x)=y \in f(A-B) \\ \implies f(A)-f(B) \subseteq f(A-B)$$

How did we get statement (*)? I didn't understand that step.

• d) is hard to parse logically. Do you want to say "$f(A)\cap f(B)=\emptyset$ for all $A,B\subseteq S$ with $A\cap B=\emptyset$"? (Similar for e) Commented Aug 23, 2019 at 7:40
• yes, that is what i meant @HagenvonEitzen
– Sam
Commented Aug 23, 2019 at 8:33

Let $$y \in f (A-B)$$. Then certainly $$y \in f(A)$$. To prove (*) we have to show that $$y \notin f(B)$$. Prove by contradiction. Suppose, if possible, $$y \in f(B)$$. Then $$y \in f(A-B)\cap f(B)$$ but this intersection is empty.
Well, if $$B\subseteq A$$ and $$f(A-B)\cap f(B)=\emptyset$$, then $$f(A-B)\subseteq f(A)-f(B)$$, since $$f(A-B)\subseteq f(A)$$.
We know that $$f(A - B) \subset f(A)$$. Then, it must also be that, as $$f(B) \subset f(A)$$, because $$f(A) = f(A - B) \cup f(B)$$. Thus, we know that the set of $$f(A-B)$$ must be contained in $$f(A)$$.
However, the statement preceding (*) tells us that $$f(A - B) \cup f(B) = \emptyset$$. Or, in other words, it tells us that $$f(A - B)$$, which is contained in $$f(A)$$, does not intersect $$f(B)$$ at all. Thus, by definition, $$f(A - B)$$ is contained within (and is thus a subset of) that subset of $$f(A)$$ which contains no points from $$f(B)$$. This subset of $$f(A)$$ is, by definition, $$f(A) - f(B)$$.